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Solve the problem.

A toy rocket is shot vertically upward from the ground. Its distance in feet from the ground at time [tex]t[/tex] seconds is given by:

[tex]s(t) = -16t^2 + 144t[/tex]

At what time or times will the rocket be 128 ft from the ground?

Answer :

We start with the given height function for the toy rocket:

[tex]$$
s(t) = -16t^2 + 144t.
$$[/tex]

We need to determine the values of [tex]$t$[/tex] for which the rocket is at [tex]$128$[/tex] feet. So, we set:

[tex]$$
-16t^2 + 144t = 128.
$$[/tex]

Step 1: Rearrange the Equation

Subtract [tex]$128$[/tex] from both sides to obtain:

[tex]$$
-16t^2 + 144t - 128 = 0.
$$[/tex]

Step 2: Simplify the Equation

Divide the entire equation by [tex]$-16$[/tex] to simplify:

[tex]$$
t^2 - 9t + 8 = 0.
$$[/tex]

Step 3: Solve the Quadratic Equation

Now, we have a standard quadratic equation:

[tex]$$
t^2 - 9t + 8 = 0.
$$[/tex]

We calculate the discriminant ([tex]$\Delta$[/tex]) using the formula:

[tex]$$
\Delta = b^2 - 4ac,
$$[/tex]

where [tex]$a = 1$[/tex], [tex]$b = -9$[/tex], and [tex]$c = 8$[/tex]. So,

[tex]$$
\Delta = (-9)^2 - 4(1)(8) = 81 - 32 = 49.
$$[/tex]

Since the discriminant is positive, there are two distinct real solutions. We use the quadratic formula:

[tex]$$
t = \frac{-b \pm \sqrt{\Delta}}{2a}.
$$[/tex]

Substitute the values:

[tex]$$
t = \frac{-(-9) \pm \sqrt{49}}{2(1)} = \frac{9 \pm 7}{2}.
$$[/tex]

This gives us the two solutions:

[tex]$$
t = \frac{9 + 7}{2} = \frac{16}{2} = 8,
$$[/tex]

and

[tex]$$
t = \frac{9 - 7}{2} = \frac{2}{2} = 1.
$$[/tex]

Step 4: Interpret the Results

The rocket reaches [tex]$128$[/tex] feet from the ground at:

[tex]$$
t = 1 \text{ second} \quad \text{and} \quad t = 8 \text{ seconds}.
$$[/tex]

Thus, the rocket is [tex]$128$[/tex] feet off the ground at [tex]$1$[/tex] second on its way up and again at [tex]$8$[/tex] seconds on its way down.

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